The solution for this problem is:
The information given lacks but nevertheless the answer is:
So, the total heat free by dissolving the solute was 1386 + 32 = 1417 J
Then, dissolving of the solute will have released -1417 J. So, per gram of Al2
(SO4)3 dissolved:
-1417 J / 25 g = -56.7 J/g
Translating that to a per mole: -56.7 J/g X 342 g/mol = -1.94X10^4 J/mol = -19
kJ/mol = this would be Delta Hsoln
Answer:
d. asteroids affected by Jupiter's gravity
Explanation:
A Trojan horse or Trojan is a type of malware that is often disguised as legitimate software. Trojans can be employed by cyber-thieves and hackers trying to gain access to users' systems. Users are typically tricked by some form of social engineering into loading and executing Trojans on their systems.
Cellular respiration releases carbon dioxide into the environment while photosynthesis pulls carbon dioxide out of the atmosphere. c:
Solve these problems like weighted averages:
The first one:
Multiply the masses (isotope numbers) by the decimal form of the percentage. Add them
0.076 (6) + 0.924 (7) = 6.924
The second one:
0.2 (10) + 0.8 (11) = 10.8
If you think about it, these answers make sense. 6.924 is much closer to 7 than to 6 (since there's a lot more lithium-7 than there is lithium-6). 10.8 is closer to 11 than to 10.
Answer: 0.151
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
![Rate=-\frac{1d[HBr]}{2dt}=+\frac{1d[H_2]}{2dt}=+\frac{1d[Br_2]}{dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7B1d%5BHBr%5D%7D%7B2dt%7D%3D%2B%5Cfrac%7B1d%5BH_2%5D%7D%7B2dt%7D%3D%2B%5Cfrac%7B1d%5BBr_2%5D%7D%7Bdt%7D)
Given: ![-\frac{d[HBr]}{dt}]=0.301](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BHBr%5D%7D%7Bdt%7D%5D%3D0.301)
Putting in the values we get:
![\frac{0.301}{2}=+\frac{1d[Br_2]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B0.301%7D%7B2%7D%3D%2B%5Cfrac%7B1d%5BBr_2%5D%7D%7Bdt%7D)
![+\frac{1d[Br_2]}{dt}=0.151](https://tex.z-dn.net/?f=%2B%5Cfrac%7B1d%5BBr_2%5D%7D%7Bdt%7D%3D0.151)
Thus the rate of appearance of 
is 0.151
